如何设置
d[i,j,i,j] = s[i,j]
使用“NumPy”并且没有for循环?
我尝试了以下内容:
l1=range(M)
l2=range(N)
d[l1,l2,l1,l2] = s[l1,l2]
答案 0 :(得分:1)
您可以使用integer array indexing(使用np.ix_
创建广播索引):
d[np.ix_(l1,l2)*2] = s[np.ix_(l1,l2)]
第一次必须复制索引(您希望[i, j, i, j]
而不是[i, j]
)这就是为什么我将tuple
返回的np.ix_
乘以2。 / p>
例如:
>>> d = np.zeros((10, 10, 10, 10), dtype=int)
>>> s = np.arange(100).reshape(10, 10)
>>> l1 = range(3)
>>> l2 = range(5)
>>> d[np.ix_(l1,l2)*2] = s[np.ix_(l1,l2)]
并确保分配了正确的值:
>>> # Assert equality for the given condition
>>> for i in l1:
... for j in l2:
... assert d[i, j, i, j] == s[i, j]
>>> # Interactive tests
>>> d[0, 0, 0, 0], s[0, 0]
(0, 0)
>>> d[1, 2, 1, 2], s[1, 2]
(12, 12)
>>> d[2, 0, 2, 0], s[2, 0]
(20, 20)
>>> d[2, 4, 2, 4], s[2, 4]
(24, 24)
答案 1 :(得分:1)
如果您考虑一下,那就与创建2D
形状(m*n, m*n)
数组并将s
中的值分配到对角线位置相同。要将最终输出设为4D
,我们最后需要重新设计。这基本上是在下面实现的 -
m,n = s.shape
d = np.zeros((m*n,m*n),dtype=s.dtype)
d.ravel()[::m*n+1] = s.ravel()
d.shape = (m,n,m,n)
运行时测试
方法 -
# @MSeifert's solution
def assign_vals_ix(s):
d = np.zeros((m, n, m, n), dtype=s.dtype)
l1 = range(m)
l2 = range(n)
d[np.ix_(l1,l2)*2] = s[np.ix_(l1,l2)]
return d
# Proposed in this post
def assign_vals(s):
m,n = s.shape
d = np.zeros((m*n,m*n),dtype=s.dtype)
d.ravel()[::m*n+1] = s.ravel()
return d.reshape(m,n,m,n)
# Using a strides based approach
def assign_vals_strides(a):
m,n = a.shape
p,q = a.strides
d = np.zeros((m,n,m,n),dtype=a.dtype)
out_strides = (q*(n*m*n+n),(m*n+1)*q)
d_view = np.lib.stride_tricks.as_strided(d, (m,n), out_strides)
d_view[:] = a
return d
计时 -
In [285]: m,n = 10,10
...: s = np.random.rand(m,n)
...: d = np.zeros((m,n,m,n))
...:
In [286]: %timeit assign_vals_ix(s)
10000 loops, best of 3: 21.3 µs per loop
In [287]: %timeit assign_vals_strides(s)
100000 loops, best of 3: 9.37 µs per loop
In [288]: %timeit assign_vals(s)
100000 loops, best of 3: 4.13 µs per loop
In [289]: m,n = 20,20
...: s = np.random.rand(m,n)
...: d = np.zeros((m,n,m,n))
In [290]: %timeit assign_vals_ix(s)
10000 loops, best of 3: 60.2 µs per loop
In [291]: %timeit assign_vals_strides(s)
10000 loops, best of 3: 41.8 µs per loop
In [292]: %timeit assign_vals(s)
10000 loops, best of 3: 35.5 µs per loop